1968 AHSME Problems/Problem 11

Problem

If an arc of $60^{\circ}$ on circle $I$ has the same length as an arc of $45^{\circ}$ on circle $II$ , the ratio of the area of circle $I$ to that of circle $II$ is:

$\text{(A) } 16:9\quad \text{(B) } 9:16\quad \text{(C) } 4:3\quad \text{(D) } 3:4\quad \text{(E) } \text{none of these}$

Solution

Let the length of the two arcs be denoted $x$ . First, we acknowledge that the length of an arc of a circle is given by the measure of that arc's central angle in radians times the radius of that circle. Note that the central angle of circle $I$ measures $\frac{\pi}{3}$ radians, and the central angle of circle $II$ measures $\frac{\pi}{4}$ radians. Let the radius of circle $I$ be $r_1$ and the radius of circle $II$ be $r_2$ . Then, we know that the arc length $x=\frac{\pi}{3}r_1=\frac{\pi}{4}r_2$ , and so $\frac{r_1}{r_2}=\frac{3}{4}$ . From this, we see that $\frac{r_1^2}{r_2^2}=\frac{\pi r_1^2}{\pi r_2^2}=\frac{[I]}{[II]}=\frac{9}{16}$ . Thus, our answer is $\fbox{(B) 9:16}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1968 AHSME Problems/Problem 11

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